2. Reciprocal
addition
Reciprocal Arithmetic
Reciprocal arithmetic is
arithmetic conjugated by the reciprocal function:
x 1/[+]
y = 1 / (1/x
+ 1/y) = xy/(x+y)
= "x with y" =
"reduction"
x 1/[-]
y = x
1/+ (-y) = xy/(y-x)
= "x without y" =
"retraction"
x 1/[*]
y = 1 /
((1/x)(1/y)) = xy
Reduction is a field under *,
with identities 1 and 1/0. Infinity is the zero of this field; and zero, in
return, is its infinity - that is, its attractor:
x
1/+ ∞ =
x
x
1/- x =
∞
x
1/+ 0 =
0
Reduction and retraction obey
these "common numerator" and "general numerator" laws:
c/a 1/+
c/b = c/(a+b)
c/a 1/-
c/b = c/(a-b)
a/b 1/+
c/d = ac / (ad+bc) =
(ad 1/+ bc) / bd
a/b 1/-
c/d = ac / (ad-bc) =
(ad 1/- bc) / bd
Reciprocal arithmetic is just
like ordinary arithmetic, but "standing on its head" - i.e. with the
roles of numerator and denominator reversed. For instance, these reciprocal
sums apply:
1
1/+ 1 = 1/2
1
1/+ 1 1/+
1
= 1/3
1
1/+ 1 1/+
1 1/+ 1 = 1/4
1
1/+ 1 1/+
1 1/+ 1 1/+ 1
= 1/5 ...
2
1/+ 2 =
1
3
1/+ 3 1/+
3
= 1
4
1/+ 4 1/+
4 1/+ 4 = 1
5
1/+ 5 1/+
5 1/+ 5 1/+ 5
= 1 ...
These formulas connect linear and
reciprocal sums:
a(a+b) 1/+
b(a+b) = ab
a(a-b) 1/-
b(a-b) = ab
a(a+1) 1/+
(a+1) = a
So a+b = c
implies that ac 1/+
bc = ab .
Therefore:
2+3=5 implies 10
1/+ 15 = 6
;
3+4=7 implies 21 1/+
28 = 12 ;
3+5=8 implies 24
1/+ 40 = 15
;
5+7=12 implies 60
1/+ 84 = 35
;
23+17=40 implies
920 1/+ 680 = 391;
and so on.
If a whole number is partitioned
into a sum of its divisors, then that implies a whole-number reciprocal
partition of unity. For instance:
6 =
1+2+3; therefore 6/6 = 6/(1+2+3) ; therefore
1 = 6 1/+
3 1/+ 2
12 =
1+2+3+6
; so 1
= 12 1/+
6 1/+ 4 1/+ 2
20 =
1+4+5+10 ; so 1 =
20 1/+ 5 1/+ 4 1/+ 2
30 =
2+3+10+15 ; so 1 =
15 1/+ 10 1/+ 3 1/+ 2
36 =
2+3+4+6+9+12 ; so 1 =
18 1/+ 12 1/+ 9 1/+ 6 1/+
4 1/+ 3
60 =
4+5+6+10+15+20 ; so
1 = 15 1/+ 12 1/+
10 1/+ 6
1/+ 4 1/+ 3
84 =
3+6+12+14+21+28 ; so 1 =
28 1/+ 14 1/+ 7 1/+
6 1/+
4 1/+ 3
90 =
3+5+9+10+15+18+30 ; so 1 =
30 1/+ 18 1/+ 10 1/+
9 1/+ 6 1/+ 5 1/+ 3
Reduction, like addition, obeys
the transposition law:
X
1/+ A =
B implies X = B
1/- A
Therefore we can solve
"reciprocal-linear" algebraic equations:
AX 1/+
B = CX 1/+ D
implies X
= (D 1/- B)
/ (A 1/- C)
The reciprocal version of the
quadratic equation:
AX2 1/+
BX 1/+ C
= ∞
has this solution:
X
= ( - 2B 1/+ ± \/{4B2 1/- AC}
) / A
The "reciprocal continued
fraction" x =
a 1/+ 1 / x
solves the equation x2 1/-
ax 1/- 1 = ∞
and therefore equals 2a 1/+ \/{4a2
1/+ 1}
Reduction connects to
multiplication via logarithmic bases:
If C = ab, then
logC(x) = loga(x) 1/+
logb(x)
If C = a/b, then
logC(x) = loga(x) 1/-
logb(x)
If C = ab, then
logC(x) = loga(x) /
b
If C = b\/a, then
logC(x) = loga(x) * b
Applications of Reduction
Reduction has a geometric
construction; the "Fence", or “Fishtail”:
* *
|\ /|
| \ / |
x | * |
y
| /|\ |
|/ |z\|
*--*--*
In this diagram the middle line,
z, has length equal to x 1/+ y. This is a construction in affine geometry; it
needs no metrics, only parallels. If we double the length of the middle line,
then we get the "reciprocal mean"
2(x 1/+ y):
"Kirkhoff's law" for
resistors involves reduction: if resistors R1 and R2 are wired in parallel, then their combined
resistance Rc equals R1 1/+ R2
.
Thus, for instance, if R1 = ∞ (a break) then Rc = ∞ 1/+
R2 = R2 ;
and if R1
= 0 (a short) then Rc
=
0 1/+ R2 =
0 .
If two lenses with focal lengths
F1 and F2 are
combined,
then the resulting lens has focal
length F1 1/+ F2 .
Given the time-distance-rate
relations D = R1T1
= R2T2 , then
D / (R1+R2) = T1
1/+ T2
D / (T1+T2) = R1
1/+ R2
D / (R1-R2) = T1
1/- T2
D / (T1-T2) =
R1 1/- R2
When rates add, times reduce, and
vice versa. "Many hands make light work".
These relations also apply to
work-rates. If Alice and Bob can do a job (say, mow the lawn) in,
respectively, T1 and T2 minutes of work, then if they work together
and their rates add, they will finish the job in T1 1/+ T2 minutes. If Bob works against Alice, then the
time will be T1 1/- T2. If Bob, that lazy bum, would take
forever to do the job, then together they finish in time T1 1/+ ∞
= T1; and if Bob, that
brilliant inventor, builds a gizmo that can do the job in no time at all, then
together they finish in time T1 1/+ 0 = 0 .
If a particle moves at velocity V1
over a distance D, then at velocity V2 over the same distance D,
then its average velocity over the distance 2D is 2(V1 1/+ V2) ; the
reciprocal mean.
If Alice's and Bob's cars are one
mile apart, and if Alice's car can drive a mile in X minutes, and Bob's car can
drive a mile in Y minutes, and if they drive toward each other, then they will
meet in (X 1/+ Y) minutes; whereas if Alice chases after Bob, then she will
overtake him in (X 1/- Y) minutes. (This is assuming that Alice's car is
faster; i.e. that X<Y.)
If Alice and Bob meet in E minutes,
and overtake in F minutes, then we obtain this reciprocal-linear system:
X 1/+
Y = E
X 1/-
Y = F
We can solve this
reciprocal-linear system by methods similar to solving a linear system; i.e. by
substitution, or by a reciprocal version of Gauss-Jordan elimination. This
yields:
X =
2 ( E 1/+ F )
; the reciprocal mean
Y =
2 ( E 1/- F )
; the reciprocal radius.
In general, the reciprocal-linear
system:
aX 1/+
bY = E
cX 1/+
dY =
F
has this reciprocal-Cramer's-rule
solution:
X =
( Ed 1/- bF )
/ ( ad 1/- bc
)
Y =
( aF 1/- Ec )
/ ( ad 1/- bc
)
That solution can be phrased in
terms of reciprocal determinants. We can also define reciprocal matrices,
vectors, etc. Reciprocal-linear algebra is just like linear algebra, except
with 1/+ standing in for +.
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